Fall 2008 Problem 2

Exercise 12.5 (Countable product of the interval)   Consider the space $X=[0,1]\times[0,1]\times\cdots$ (the countably-infinite product of $[0,1]$ with the product topology). An element of $X$ may be thought of as a sequence $\{x_n\}_{n=1}^\infty$ with each $x_n\in[0,1]$. Show that the function from $X$ to $\mathbb{R}$ defined by

$$\{x_n\} \mapsto \sum_{n=1}^\infty 2^{-n}x_n$$ (12.39)

is continuous.

Proof. Let $U\subseteq\mathbb{R}$ be an open set. Select $f(y)\in B_\epsilon(p)\subseteq U$. Let $\epsilon'=\epsilon-\vert f(y)-p\vert$. By a metric space argument, $B_{\epsilon'}(f(y)) \subseteq B_\epsilon(p)\subseteq U$. Select $N$ such that

$$\sum_{n=N}^\infty \frac{2}{2^n} < \epsilon'/2.$$ (12.40)

Define a basic open set in $X$

$$V = \prod_{n=1}^{N-1} B_{\epsilon'/2}(y_n) \times \prod_{n=N}^\infty [0,1].$$ (12.41)

Certainly $y\in V$, and we show that $V$ is interior to the inverse image. Let $\widetilde{y}\in V$. Then

$$\vert f(y)-f(\widetilde{y})\vert = \left\vert \sum_{n=1}^\infty \frac{y_n}{2^n} - \frac{\widetilde{y_n}}{2^n} \right\vert$$ (12.42)

$$\leq \sum_{n=1}^{N-1}\frac{\vert y_n-\widetilde{y_n}\vert}{2^n} + \sum_{n=N}^\infty \frac{\vert y_n-\widetilde{y_n}\vert}{2^n}$$ (12.43)

$$\leq \sum_{n=1}^{N-1}\frac{\epsilon'/2}{2^n} + \sum_{n=N}^\infty\frac{2}{2^n}$$ (12.44)

$$\leq \epsilon'/2 + \epsilon'/2$$ (12.45)

$$\leq \epsilon'$$ (12.46)

Therefore, $f(\widetilde{y})\in B_{\epsilon'}(f(y))\subseteq U$, indicating $\widetilde{y}\in f^{-1}(U)$. $\qedsymbol$